问题:从字符串列表的元素中删除结尾的换行符

我必须采用以下形式的大量单词:

['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']

然后使用strip功能,将其转换为:

['this', 'is', 'a', 'list', 'of', 'words']

我以为我写的东西行得通,但是我不断收到错误消息:

“’list’对象没有属性’strip’”

这是我尝试过的代码:

strip_list = []
for lengths in range(1,20):
    strip_list.append(0) #longest word in the text file is 20 characters long
for a in lines:
    strip_list.append(lines[a].strip())

I have to take a large list of words in the form:

['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']

and then using the strip function, turn it into:

['this', 'is', 'a', 'list', 'of', 'words']

I thought that what I had written would work, but I keep getting an error saying:

“‘list’ object has no attribute ‘strip'”

Here is the code that I tried:

strip_list = []
for lengths in range(1,20):
    strip_list.append(0) #longest word in the text file is 20 characters long
for a in lines:
    strip_list.append(lines[a].strip())

回答 0

>>> my_list = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
>>> map(str.strip, my_list)
['this', 'is', 'a', 'list', 'of', 'words']
>>> my_list = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
>>> map(str.strip, my_list)
['this', 'is', 'a', 'list', 'of', 'words']

回答 1

清单理解力? [x.strip() for x in lst]

list comprehension? [x.strip() for x in lst]


回答 2

您可以使用列表推导

strip_list = [item.strip() for item in lines]

map功能:

# with a lambda
strip_list = map(lambda it: it.strip(), lines)

# without a lambda
strip_list = map(str.strip, lines)

You can use lists comprehensions:

strip_list = [item.strip() for item in lines]

Or the map function:

# with a lambda
strip_list = map(lambda it: it.strip(), lines)

# without a lambda
strip_list = map(str.strip, lines)

回答 3

这可以使用PEP 202中定义的列表理解来完成

[w.strip() for w in  ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']]

This can be done using list comprehensions as defined in PEP 202

[w.strip() for w in  ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']]

回答 4

所有其他答案,主要是关于列表理解的,都很棒。但是只是为了解释您的错误:

strip_list = []
for lengths in range(1,20):
    strip_list.append(0) #longest word in the text file is 20 characters long
for a in lines:
    strip_list.append(lines[a].strip())

a是您列表的成员,而不是索引。您可以这样写:

[...]
for a in lines:
    strip_list.append(a.strip())

另一个重要的评论:您可以通过以下方式创建一个空列表:

strip_list = [0] * 20

但这不是那么有用,因为可以.append 内容追加到列表中。在您的情况下,创建带有默认值的列表是没有用的,因为在附加剥离字符串时,将逐项构建该列表。

因此,您的代码应类似于:

strip_list = []
for a in lines:
    strip_list.append(a.strip())

但是,可以肯定的是,最好的选择就是这个,因为这是完全一样的:

stripped = [line.strip() for line in lines]

如果您遇到的不仅仅是a复杂的事情.strip,请将其放在函数中并执行相同的操作。这是使用列表最易读的方式。

All other answers, and mainly about list comprehension, are great. But just to explain your error:

strip_list = []
for lengths in range(1,20):
    strip_list.append(0) #longest word in the text file is 20 characters long
for a in lines:
    strip_list.append(lines[a].strip())

a is a member of your list, not an index. What you could write is this:

[...]
for a in lines:
    strip_list.append(a.strip())

Another important comment: you can create an empty list this way:

strip_list = [0] * 20

But this is not so useful, as .append appends stuff to your list. In your case, it’s not useful to create a list with defaut values, as you’ll build it item per item when appending stripped strings.

So your code should be like:

strip_list = []
for a in lines:
    strip_list.append(a.strip())

But, for sure, the best one is this one, as this is exactly the same thing:

stripped = [line.strip() for line in lines]

In case you have something more complicated than just a .strip, put this in a function, and do the same. That’s the most readable way to work with lists.


回答 5

如果您只需要删除结尾的空格,则可以使用str.rstrip(),它的效率应比str.strip()

>>> lst = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
>>> [x.rstrip() for x in lst]
['this', 'is', 'a', 'list', 'of', 'words']
>>> list(map(str.rstrip, lst))
['this', 'is', 'a', 'list', 'of', 'words']

If you need to remove just trailing whitespace, you could use str.rstrip(), which should be slightly more efficient than str.strip():

>>> lst = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
>>> [x.rstrip() for x in lst]
['this', 'is', 'a', 'list', 'of', 'words']
>>> list(map(str.rstrip, lst))
['this', 'is', 'a', 'list', 'of', 'words']

回答 6

my_list = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
print([l.strip() for l in my_list])

输出:

['this', 'is', 'a', 'list', 'of', 'words']
my_list = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
print([l.strip() for l in my_list])

Output:

['this', 'is', 'a', 'list', 'of', 'words']

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